Modeling Snow Crystal Growth I: Rigorous Results

Transcription

Modeling Snow Crystal Growth I:
Rigorous Results for Packard’s Digital Snowflakes
Janko Gravner and David Griffeath
CONTENTS
1. Introduction
2. Extreme Boundary Dynamics
3. Densities for Exactly Solvable Rules
4. Density of Hex 1 and Its Cousins
5. Density of Hex 1456 and Hex 146
6. Proof of Theorem 1.1
7. Macroscopic Dynamics
8. Thickness and Related Issues
9. Exact Solvability
Acknowledgments
References
Digital snowflakes are solidifying cellular automata on the triangular lattice with the property that a site having exactly one
occupied neighbor always becomes occupied at the next time
step. We demonstrate that each such rule fills the lattice with
an asymptotic density that is independent of the initial finite set.
There are some cases in which this density can be computed
exactly, and others in which it can only be approximated. We
also characterize when the final occupied set comes within a
uniformly bounded distance of every lattice point. Other issues
addressed include macroscopic dynamics and exact solvability.
1.
2000 AMS Subject Classification: Primary 37B15;
Secondary 68Q80, 11B05, 60K05
Keywords: Asymptotic density, cellular automaton, exact
solvability, growth model, macroscopic dynamics, thickness
INTRODUCTION
Six-sided ice crystals that fall to earth in ideal winter conditions, commonly known as snowflakes, have fascinated
scientists for centuries. They exhibit a seemingly endless variety of shape and structure, often dendritic and
strangely botanical, yet highly symmetric and mathematical in their designs. To this day, snowflake growth from
molecular scales, with its tension between disorder and
pattern formation, remains mysterious in many respects.
Study of snowflakes dates back at least to the sixteenth
century [Magnus 55], and includes early contributions
from such scientific giants as Kepler [Kepler 66], Hooke
[Hooke 03], and Descartes [Descartes 37]. With the advent of cameras came the first snow crystal album: more
than five thousand photos collected by W. Bentley beginning in 1885 [Bentley and Humphreys 62]. Although
rather few people have ever seen such crystals with their
own eyes, Bentley’s images helped establish snowflake
designs, simplified and idealized, as universal icons for
wintertime. The most significant scientific advances of
the past century were due to Nakaya [Nakaya 54] in the
1930s, who classified natural crystals into dozens of types,
first grew synthetic crystals in the laboratory, and discovered an elaborate, still perplexing morphology diagram,
c A K Peters, Ltd.
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Experimental Mathematics 15:4, page 421
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Experimental Mathematics, Vol. 15 (2006), No. 4
which predicts the predominant type of snowflake arising at any given temperature and supersaturation level.
(In particular, the familiar essentially two-dimensional
crystals arise only for certain parameter values; in other
conditions, columnar “needles” form.) All this and much
more is explained superbly in a recent popular account by
Libbrecht and Rasmussen [Libbrecht and Rasmussen 03],
which also contains a great many state-of-the-art photographs of breathtaking beauty. There is a companion
web site [Libbrecht 06]; see also [Libbrecht 05] for a current scholarly review.
Over the past century, geometric structures inspired
by snow crystals have begun to adorn the world of
mathematics. Most celebrated is the Koch snowflake
[King 64], introduced by H. von Koch [von Koch 04] in
1904. One of the earliest known fractals, a closed curve
with Hausdorff dimension 2 log 2/log 3, is obtained in
the limit by starting from an equilateral triangle (with
• markers at its vertices) and repeatedly applying the
substitution scheme in the diagram below to each piece
between markers:
More recent variations on Koch’s construction include
Gosper’s flowsnake [Gardner 76] and the pentaflake
[Dixon 91]. While none of these designs resembles a real
snow crystal to any great extent, their blend of elementary polygonal shapes with infinitely fine branching detail
evokes the same iconography as Bentley’s album.
The building blocks for snowflakes are hexagonally arranged molecules of natural ice (Ih). Just how the elaborate designs emerge as water vapor freezes is still poorly
understood. Only very recently have a few rudimentary movies of synthetic crystal growth been produced
[Libbrecht and Rasmussen 03, p. 57]. The solidification
process involves complex physical chemistry of diffusionlimited aggregation and attachment kinetics. Theoretical
research and mathematical modeling to date have mainly
focused on the evolution of dendrite tips. See [Gleick
87, pp. 309–314] for a popular account of the challenges,
[Meakin 98] and [Pimpinelli and Villain 99] for recent
scholarly monographs, and [Adam 05] for a current review.
In 1984, Packard [Packard 86] introduced a supremely
simple cellular automaton (CA) model for crystal solidification. On a honeycomb lattice of hexagonal cells,
start with a single “seed” cell of ice surrounded by vapor. At each subsequent discrete-time update, any vapor cell neighboring the requisite number of frozen cells
turns to ice. Since real snowflake growth favors the tips of
the crystal, Packard proposed that exactly one occupied
(frozen) neighbor should cause solidification, but exactly
two should not. Thus, in one of his digital snowflakes a
site joins the crystal if and only if it has exactly one occupied neighbor, while in another it joins if the number of
occupied neighbors is odd. In the present paper we will
refer to these rules as Hex 1 and Hex 135 , respectively.
Packard’s snowflake automata have been widely publicized since the 1980s to illustrate how very simple mathematical algorithms can emulate complex natural phenomena. A multicolor image of Hex 135 occupies nearly
all of page 189 in Wolfram’s 1984 article [Wolfram 84a],
and the same graphic is reproduced as the first color
plate of Steven Levy’s 1992 book [Levy 92]. More recently, the first 30 updates of Hex 1 are illustrated on
page 371 of [Wolfram 02]. The central tenet of [Wolfram
84a], already familiar from the established universality
of Conway’s game of life [Berlekamp et al. 04], was that
“Simulation by computer may be the only way to predict how certain complicated systems evolve.” Implicit
in this perspective is the inadequacy of mathematics to
analyze complexity. In the discussion of Hex 135 and the
corresponding caption, he writes,
Snowflakes grown in a computer experiment
from a single frozen cell according to this rule
show intricate treelike patterns, which bear
a close resemblance to real snowflakes.. . . The
only practical way to generate the pattern is by
computer simulation.
Levy’s account reiterates the claimed verisimilitude:
An elementary schoolchild could look at any
of the gorgeous pictures of computer screens in
Packard’s collection and instantly identify it as
a snowflake.
So how do these digital snowflakes evolve? The left
frame of Figure 1 shows a representative snapshot of Hex
134 after 218 updates starting from a single occupied
cell. (The graphic in [Wolfram 84a] and [Levy 92] is
quite similar.) Letting At denote the crystal at time t,
started from A0 = {0}, it turns out that A2n occupies,
with a certain density, the hexagon of lattice cells within
2n steps of the origin for each n, but that shapes with
apparently fractal boundary arise in the limit along intermediate subsequences of the form tn = a2n when a
is not a dyadic rational. For instance, the limit shape
for Hex 1 and Hex 135 along the a = 13 subsequence is
exactly the Koch-type snowflake starting from a regular
Gravner and Griffeath: Modeling Snow Crystal Growth I: Rigorous Results for Packard’s Digital Snowflakes
hexagon (with • markers now in the middles of edges)
and based on the substitution scheme shown in the diagram below, applied to each nonstraight segment between
markers:
Here, the top and the bottom choices apply to a concave and a convex vertex, respectively. (A vertex is convex (concave) if one makes a right (left) turn at it while
moving counterclockwise on the curve.)
In other words, the crystal oscillates between hexagonal shapes and other familiar mathematical forms with
increasingly complex boundary. To illustrate the scientific insight offered by such simple CA rules, Wolfram
makes the following intriguing prediction based on his
time trace of Hex 1 up to time 30 [Wolfram 02]:
For example, one expects that during the
growth of a particular snowflake there should
be alternation between tree-like and faceted
shapes, as new branches grow but then collide with each other. And if one looks at real
snowflakes, there is every indication that this
is exactly what happens. And in fact, in general the simple cellular automaton shown above
seems remarkably successful at reproducing all
sorts of obvious features of snowflake growth.
We intend to address the level of realism of digital
snowflakes and other lattice models for ice crystal growth
in a sequel to this paper [Gravner and Griffeath 06].
Here our goal is a rigorous study of the complete family of Packard rules. For instance, we will show that each
crystal fills the lattice with a characteristic asymptotic
density, independent of the finite initial seed. We also
identify a subclass of exactly solvable rules for which the
density is a computable rational number (e.g., Hex 135
and Hex 134 have densities 56 and 21
22 , respectively), and
a complementary class with inherently more computational complexity of the limit state A∞ . Ultimately, our
story here has two main morals:
• As CA dynamics go, Packard’s digital snowflakes are
not very complex. The patterns they trace are not
periodic, but nearly so, and in some cases exactly
described by a finite recursion. Consequently, they
are quite amenable to mathematical methods.
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• While computer visualization and empirical calculation are indispensable tools, subtle properties of
these dynamics cannot possibly be gleaned from simulations alone. Deductive reasoning plays a fundamental role in the analysis.
Turning to formalities, our basic setup features solidification CA on the triangular lattice T (to reflect the
arrangement of water molecules in ice crystals).1 For
notational and computational convenience, we represent
T by Z2 , with the neighborhood N of (0, 0) consisting
of itself and the six sites (±1, 0), (0, ±1), and ±(1, 1).
The neighborhood of an arbitrary x is then x + N . This
representation is handy because integers are much more
familiar than Cayley graph representations of T. On the
other hand, some symmetries are not so easy to spot.
(Figures 1 and 8 are the only ones in which dynamics on
T are depicted.) The CA simulator MCell [Wójtowicz
05] is ideally suited for empirical investigation of digital
snowflakes, since its Weighted Life rule set supports this
Z2 embedding. Indeed, MCell was a crucial resource during the early stages of our work, and any conscientious
reader of this paper will surely need to enlist its aid, or
that of some similar program.
We denote by At ⊂ Z2 the set of occupied sites at
time t. Often, sites in At are called 1’s, and sites in
Act are called 0’s. The set At grows in discrete time
t = 0, 1, 2, . . . . That is, At ⊂ At+1 ; such CA are called
solidifying. Whether x ∈
/ At belongs to At+1 depends
only on the number of sites it sees in At , that is, on
|(x + N ) ∩ At |. Thus, the rule is given by a function π : {1, 2, 3, 4, 5, 6} → {0, 1} such that for x ∈
/ At ,
π(|(x + N ) ∩ At |) = 1 iff x ∈ At+1 . As above, we specify π by listing all n for which π(n) = 1. Our canonical
choice of the initial set is A0 = {0}, although we will also
study the dynamics started from an arbitrary finite set,
and use assorted infinite initial sets as props. Note that
for any solidification dynamics and every A0 , the final set
A∞ exists as a sitewise limit of At , and A∞ = ∪t≥0 At .
Our basic assumption on π, reflecting the fact that
tips of a growing snow crystal are favored, is that π(1) =
1. We call solidifying CA with this property digital
snowflakes. It is not clear what else should be assumed
about π, so we propose to catalog all possible behaviors
of such automata.
There are 32 candidates for digital snowflakes. Of
these, 16 are trivial: when π(2) = 1, the appropriate ver1 We have chosen to conform to accepted mathematical terminology, although T is often called the hexagonal lattice in the popular
and scientific literature.
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FIGURE 1. The occupied set of Hex 134 at time 218, started from {0}. Let τx = inf{t : x ∈ At } be the time x is occupied
and d the distance on the triangular lattice. The two frames depict periodically shaded contours of constant τx (left) and
constant σx = τx − d(x, 0) (right, where only the 0 contour {x : σx = 0} is black).
sion of the extreme boundary dynamics (see Section 2) is
very easy to analyze, and causes the dynamics to grow a
full hexagon from a singleton, while from any other finite
seed they grow a full hexagon apart from a region at finite distance from the rays defined by the extreme points
of the hexagon. (Many real snowflakes also grow as expanding hexagons, but there is no mystery here.) From
now on, we refer to only the remaining 16 rules, i.e., to
those with π(1) = 1 and π(2) = 0, as digital snowflakes.
Next, let us explain our basic approach and summarize
our asymptotic density results.
The right frame of Figure 1 highlights key structural
features that underlie our analysis. Note first that any
digital snowflake advances at speed 1 along the axes of
the lattice, since the closest unoccupied site in these directions always has only one occupied neighbor. Within
each of six wedges formed by the axes, the black cells are
those that solidify at the edge of the light cone, i.e., at
the maximum speed of propagation allowed by a nearestneighbor rule. This process induces symmetric copies of
the space-time pattern of a one-dimensional CA. Because
π(1) = 1 and π(2) = 0, this is the additive xor rule, arguably the most familiar of all cellular automata. Consequently, the black cells form discrete versions of a famous
fractal known as the Sierpiński triangle. Of course, digital snowflakes continue to solidify after the edge of the
light cone passes, as seen in the gray portions of the right
frame of Figure 1. But the Sierpiński lattice effectively
divides the crystal into independent finite regions with
all 1 boundary conditions, within which subsequent dynamics evolve.
To summarize, Packard’s snowflakes enjoy three important properties (to be precisely described in Section 2):
• Starting from a single occupied cell, the light cone
CA forms an impenetrable web of occupied sites that
divides further solidification into independent finite
domains with simple boundary conditions.
• Boundary effects within each domain are controlled.
• The light cone CA is additive, so the web from a
general finite seed is representable as a superposition
of webs from each of its individual cells.
The first two properties ensure a recursive representation of the dynamics, while the last is crucial for the
asymptotic density’s independence of A0 .
The delicacy of our results is conveyed effectively by
comparison to analogous solidification on Z2 with range1 Box neighborhood consisting of a central cell and its
eight nearest neighbors: (0, 0), (0, ±1), (±1, 0), (±1, ±1).
There are 128 such rules with π(1) = 1; see [Griffeath 06]
for a brief introduction and colorful graphics of Box 1,
Box 157, Box 1357 , and Box 136 crystals. Although
snowflake-like recursive carpets emerge in a great many
cases, any and all of the three properties above may fail.
For instance, we will see in Section 6 that the density of
Box 1 , provided it exists at all, can depend on the initial
seed. Also, for the “odd” rule, Box 1357 , the light-cone
web “leaks,” and growth is apparently chaotic. Although
there are many fascinating problems connected with the
Box neighborhood, and exact computations are feasible
in some cases, there is no hope of a complete analysis as
in the present Hex setting.
Fix a set S ⊂ Z2 . Let µ be 2 times the counting
measure on · S. We say that S has asymptotic density
ρ if µ converges to ρ · λ as → 0. Here λ is Lebesgue
measure on R2 , and the convergence holds in the usual
sense:
(1–1)
f dµ → ρ · f dλ
425
sites on a side, one would still be led to the conclusion
that A∞ = Z2 . In fact, the four dynamics are all distinct
eventually and presumably have different densities less
than one.
In contrast to density results, which are macroscopic
in nature, our next result addresses the most basic microscopic properties of final configurations. Call a set
S ⊂ Z2 thick if
sup{d(x, S) : x ∈ Z2 } < ∞.
for any f ∈ Cc (R2 ).
Theorem 1.1. To each of the 16 digital snowflakes there
corresponds a ρ ∈ (0, 1], the asymptotic density of A∞ ,
that is independent of the finite seed A0 .
We will index the densities by our notation for the respective rules, and give more information on their values
in the next theorem.
Theorem 1.2. The densities are exactly computable in
eight cases:
ρ13 = ρ135 = 5/6 ≈ 0.8333,
ρ134 = ρ1345 = 21/22 ≈ 0.9545,
ρ136 = ρ1356 = ρ1346 = ρ13456 = 1.
In six other cases, one can estimate, within ±0.0008,
ρ1 ≈ 0.6353,
ρ14 , ρ145 ≈ 0.9689,
ρ15 ≈ 0.8026,
ρ16 ≈ 0.7396,
ρ156 ≈ 0.9378.
Finally,
ρ146 ∈ (0.995, 1), ρ1456 ∈ (0.9999994, 1).
Perhaps surprisingly, ρ14 > ρ134 , testimony to the fundamentally nonmonotone nature of these rules.
We refer to the first eight rules in Theorem 1.2 as
exactly solvable. In Section 9, we will develop a rigorous
foundation for this terminology.
It is tempting to conjecture that ρ14 = ρ145 and ρ146 =
ρ1456 since the two dynamics of each pair are identical
starting from A0 = {0} on finite arrays up to 500 × 500
in size. This question remains open, but one should resist
such empirical conclusions. For instance, as we shall see
later, observing Hex 1456 from A0 = {0} on even the
world’s most extensive graphics array, with millions of
Here, d is distance in any chosen norm, say ·
∞ . For
snowflakes with density between 0 and 1, thickness of A∞
and Ac∞ is one rough notion of an almost-periodic final
state. In the following theorem, A0 is assumed to be an
arbitrary finite set.
Theorem 1.3. The eight exactly solvable rules have the
following properties:
(1) The final set A∞ is always thick.
(2) Hex 13456 always has A∞ = Z2 . For the other rules
with density 1, there exist initial conditions for which
A∞ contains infinitely many 0’s.
(3) Ac∞ is always thick for rules with density less than 1,
and never thick for those with density 1.
For the eight rules that are not exactly solvable, A∞ is
never thick, and Hex 1 always has thick Ac∞ .
It is an intriguing open question whether Ac∞ is thick
for the seven rules not covered by Theorem 1.3. We suspect that it is for all of them, but have no argument.
The rest of the paper is organized as follows. Preliminaries in Section 2 describe precisely various structural
properties of the additive web that decomposes a digital
snowflake into independent regions of finite size. Slight
variations in how these regions solidify are identified case
by case. Sections 3 and 4 then detail the first eight density calculations of Theorem 1.2 by deriving and solving
recursions for |A2n −1 |. Section 3 handles exactly solvable cases, first the simplest rules: 13 , 135 , 136 , and
1356 , and then those obeying slightly more complicated
dynamics: 134 , 1345 , 1346 , and 13456 . For all these
snowflakes, the limit of |A2n −1 |/(3 · 4n ) is evaluated explicitly. Next, Sections 4 and 5 develop and analyze corresponding recursions for Hex 1 and the other seven rules
that are not exactly solvable. Now, due to certain messy
interactions, existence of the normalized limit of occupied
cells is established by a novel application of the renewal
426
theorem, but this density is implicit and can only be approximated numerically. Section 4 handles six densities
that we are able to estimate within 0.0008. The cases
146 and 1456 in Section 5 require a different rescaling
argument for the upper bound, since their densities are
extremely close to, but less than, 1.
In Section 6 we complete the proof of Theorem 1.1 by
showing that A∞ has an asymptotic density ρ starting
from {0} in the formal sense of (1–1), that ρ agrees with
the corresponding value obtained in Sections 3–6, and
that the same asymptotic density occurs when the initial
seed is an arbitrary finite set A0 . (A technicality for rules
with π(3) = 0 is also handled at the end of this section.)
Then, Section 7 introduces two distinct rule-dependent
macroscopic dynamics Sa for the limit of 2−n Atn , where
Theorem 7.1 thereby extends to
tn = a · 2n .
general a the already-mentioned substitution-scheme
limit for Hex 1 and Hex 135 in the case a = 13 . A more
sophisticated approach is required to handle both holes
formed by colliding branches of the snowflake and the
case of irrational a. Examples are given to illustrate
the exotic dependence on a of the Hausdorff dimension of the boundary of Sa . Finally, Sections 8 and 9
address thickness (Theorem 1.3) and exact solvability,
respectively. Theorems 1.1–1.3 yield a natural and precise division of digital snowflakes into two complexity
classes (Section 9). This distinction, based on the notion of automaticity of the final set, is potentially widely
applicable in the computational theory of cellular automata.
2.
EXTREME BOUNDARY DYNAMICS
Our basic tool is the additive dynamics Tn , also referred
to as xor , addition mod 2, or rule 90 (see, e.g., [Willson
84]). Among several equivalent definitions, we choose the
following. We declare the neighborhood of 0 to consist of
(−1, 0) and (−1, −1), and the additive rule to be exactly
one solidification dynamics with this two-point neighborhood, i.e., according to this rule a site changes its state
to 1 iff exactly one of x + (−1, 0) and x + (−1, −1) is in
state 1. The initial set T0 will always be a subset of the
y-axis, our default choice being the singleton {0}. If we
want to emphasize that T0 = A, we use the notation TnA .
Many properties of this rule are well known and easy to
check. Nevertheless, we will explain them briefly as they
are used.
Observe first that with the canonical choice T0 = {0},
{(x, 0) : 0 ≤ x ≤ n} ∪ {(x, x) : 0 ≤ x ≤ n}
⊂ Tn ⊂ {(x, y) : 0 ≤ y ≤ x ≤ n}.
(2–1)
One can also prove by induction that
T2n −1 ∩ {(x, y) : x = 2n − 1}
= {(2n − 1, y) : 0 ≤ y ≤ 2n − 1}
(2–2)
and
T2n ∩ {(x, y) : x = 2n } = {(2n , 0), (2n , 2n )},
(2–3)
since the two “buds” at time 2n create two versions of
the dynamics that do not interact through time 2n+1 − 1.
The name of the dynamics stems from the fact that it
preserves exclusive union: TnA xor B = TnA xor TnB . This is
immediate for n = 1, and then again follows by induction.
It is helpful to consult Figures 2–4 while reading the
remainder of this section. The darker sites in those figures form initial conditions (the reasons for which will be
explained later). For growth from a single seed at the
origin, the lowest row (two rows) in the top three frames
of Figures 2 and 3 (Figure 4) should be deleted, and then
the origin placed at the leftmost lowest site. Also, time
should be diminished by 1 in Figures 2 and 3.
The relevance of additive dynamics to digital
snowflakes becomes apparent when we note that if A0
does not include any site to the right of the y-axis,
An ∩ {(x, y) : x = n} = TnA0 ∩y-axis ∩ {(x, y) : x = n}
(2–4)
in any of our 16 dynamics. For a general solidification CA, the light cone of a set L0 is the set Ln =
L0 + N + N + · · · + N , where N is repeated n times,
i.e., the set of points that can possibly be influenced by
L0 at time n. In particular, if L0 = A0 , which we assume from now on, An ⊂ Ln . The extreme boundary
comprises sites y ∈ Ln with (x + N ) ∩ Lcn = ∅. Then
(2–4) means that our rules perform six copies of the additive dynamics (appropriately mapped) at their extreme
boundaries, and create an “additive web” consisting of
black sites x with σx = 0 in the right frame of Figure
1. We will call this the extreme boundary dynamics, and
the sites so created primary boundaries. Also note that
properties (2–1) and(2–2) dictate impenetrable boundaries for our rules. For example, if A0 = {0}, then after
time 2n − 1 the dynamics create a triangle of 1’s given by
(2–1) and (2–2), and after that time the dynamics inside
and outside the triangle are independent. Even (2–1) itself, since it extends all the way to the extreme boundary,
separates the dynamics into six independent wedges. For
many purposes, then, it will suffice to look at one of these
wedges, typically the one in the first quadrant and below
the line y = x.
Digital snowflakes do more than add sites at their extreme boundaries. At time 2n , for example, the two buds
in (2–3) do not merely spread into two new triangles outlined by the additive dynamics; they also grow into the
empty triangle between them. To be more precise, in
the eight rules with π(3) = 1, there is a pair of occupied
sites (2n , 1), (2n , 2n − 1) ∈ A2n +1 ; in the remaining eight
cases, (2n + 1, 2), (2n + 1, 2n − 1) ∈ A2n +2 . In fact, it
is more convenient to interpret the initial buds as being
at sites below and above these two; they are marked •
in the top left configurations of Figures 2 and 4. As we
explain below, these two buds generate their own secondary extreme boundary dynamics (spreading into the
two smaller triangles outlined in the figures) until they
collide.
Assume first that π(3) = 1. Then the two buds generate exactly (appropriately rotated and deformed) additive dynamics at their extreme boundaries. This is because one of the two endpoints (sites generating the leftmost set in (2–1)) sees three occupied sites, two of which
are contributed by the boundary conditions (occupied
sites in primary boundaries), while the other endpoint is
shared by the neighboring additive dynamics. Therefore,
the said two extra buds generate two copies of additive
extreme boundary dynamics (in the smaller triangles),
which at the time 2n + 2n−1 − 1 generate an occupied
secondary row and diagonal of length 2n−1 , separated by
a 0 at (2n , 2n−1 ). This 0, let us call it the four-site, sees
four occupied sites.
If π(4) = 0, the four-site (marked by ◦ in the top middle of Figure 2) will not get occupied immediately, and
the vertex of the wedge between the row and the diagonal
of 1’s is “dead.” The two secondary boundaries, together
with the primary ones {(x, 2n ), (x, x − 2n ) : 2n + 2n−1 ≤
x ≤ 2n+1 −1}, form a “hole,” which is invaded by dynamics that emanate from two one-buds at (2n + 2n−1 , 2n−1 )
and (2n + 2n−1 , 2n ), created at time 2n + 2n−1 (each
marked by • in the top middle of Figure 2). This hole is
in turn divided into two smaller holes (by secondary 1’s)
at time 2n + 2n−1 + 2n−2 − 1, etc. This hole-filling mechanism is illustrated in the bottom of Figure 2, where the
two descendant holes are outlined in the first two frames.
It is important to note that the parallelogram hole in
the top row of this figure is equivalent, modulo boundary
corrections, to the small square hole with darker shaded
1 boundary conditions. (Match the marked first two occupied sites with the same marks in the parallelogram
hole.) This consequence of symmetries of T will be exploited throughout. Armed with these observations, we
427
will begin our study of the Hex 13 rule in the next section.
We emphasize that some of the sites that we use as initial conditions are not present initially; the hole “frames”
are typical examples. Indeed, some of these initial sites
may never be created, as in the Hex 1 case. However,
they are very convenient for definitions and for symmetry considerations. In each case, it is straightforward to
verify that the dynamics behaves as if these sites were
present initially.
On the other hand, if π(3) = π(4) = 1, then the foursite (now marked • in Figure 3) becomes occupied at time
2n +2n−1 , and the resulting “live” vertex gives rise to another secondary dynamics in the hole. Three secondary
boundary dynamics originating at the four-site and the
two one-buds all collide at time 2n + 2n−1 + 2n−2 − 1 to
create a triangular hole (as in the middle top of Figure
3). The remaining two secondary boundary dynamics
collide at the same time and create a smaller quadrilateral hole. This one, however, is of a type different from
that of the original, since only two of its vertices are live.
This mechanism is iterated, as illustrated by a larger hole
example in Figure 3. Our analysis of Hex 134 dynamics will therefore require three types of hole dynamics,
generated by different initial conditions.
The situation is again different when π(3) = 0. Now
the two secondary buds are a little off center (by one
site, to be precise). They still generate additive boundary dynamics, but the final interaction inside the resulting hole generates two holes of different sizes, and
each successive generation of holes has one of a smaller
size. We give a more-precise description for the Hex 1
rule; others are similar. In this case, all points (2n , y),
1 ≤ y ≤ 2n − 1, see at least two occupied sites to their
left at time 2n − 1 (hence thereafter) and thus will never
get occupied. The two secondary buds collide at time
2n + 2n−1 , but the secondary row and diagonal (created
at time 2n + 2n−1 ) are now not separated. Nevertheless, the 0 at (2n + 2, 2n−1 + 1) (marked ◦ in Figure 4)
sees four occupied points, so it never gets occupied, and
the resulting vertex of the wedge is dead.2 The two secondary boundaries, together with primary ones, create a
hole. This hole starts being filled by buds that appear
at time 2n + 2n−1 + 2 at (2n + 2n−1 + 1, 2n−1 + 1) and
(2n +2n−1 +1, 2n ). The fact that these buds are off-center
has two consequences. The first is minor and technical: it
is necessary to start the analysis with a basic wedge and
holes that incorporate the buds in their initial conditions.
2 In rules with π(4) = 1, such as Hex 14 , this 0 becomes a 1, but
then growth from the vertex stops.
428
FIGURE 2. Hex 13 : Basic wedge dynamics (top) at times 18, 24, and 32; final configuration of a hole of size 10; size-34-hole
dynamics at times 16, 24, and 32.
The second fact is crucial for the analysis of these rules.
Namely, the two holes that result when the additive dynamics from two secondary buds collide are of unequal
size, their sizes differing by exactly 2. This in turn creates holes of a larger and larger variety of sizes (as shown
in the two bottom frames of Figure 4, where the secondgeneration descendant holes are also outlined, since they
evolve slightly out of phase and are thus difficult to identify). That the interaction in smaller holes still creates
impenetrable boundaries and dead wedge vertices is guaranteed by the following lemma.
Lemma 2.1. Assume that the initial configuration consists
of two points, A0 = {(0, 0), (2n − k, 2n − k), 0 < k ≤
2n−1 }. For any digital snowflake,
{(y, 2n − 1) : 0 ≤ y ≤ 2n−1 − k + 1} ⊂ A2n −1 .
Proof: This is a simple consequence of the speed of light.
Namely, consider the light cone of the point (2n − k, 2n −
k) at time t. Outside this light cone, the dynamics started
from {(0, 0)} and the one started from A0 agree through
time t. Apply this observation at time t = 2n − 1.
Assume first that the initial occupied set is the origin,
A0 = {0}, and define
ρ13 = lim
n→∞
|A2n −1 |
.
3 · 4n
We will see shortly that the limit exists, and that
bn
,
n→∞ 4n
ρ13 = lim
where bn are defined by certain wedge dynamics. Namely,
run the dynamics An inside the first quadrant with one
boundary condition on the axes. Let
Bn = {(x, y) : 1 ≤ x ≤ 2n , 1 ≤ y ≤ 2n }.
Then bn is the size of the final occupied set inside Bn ,
bn = |A∞ ∩ Bn | = |A2n ∩ Bn |.
The second equality, i.e., the fact that these dynamics do
not occupy any site on Bn past time 2n , can be proved
by induction along the lines of arguments given below.
Smaller wedge numbers are obtained with the same
initial condition as for bn but counted only inside
Wn = {(x, y) : 1 ≤ x ≤ 2n , 1 ≤ y ≤ x}
and defined by
3.
DENSITIES FOR EXACTLY SOLVABLE RULES
We begin with the Hex 13 rule, since it represents the
simpler of two exactly solvable cases with nontrivial
density.
wn = |A∞ ∩ Wn |.
The dynamics inside this wedge (with given initial conditions) will be called the basic wedge dynamics. See the
top row of Figure 2. Note that bn = 2wn − 2n .
By direct enumeration, b0 = 1, b1 = 4, b2 = 14, b3 =
54, etc. Moreover, since A2n −1 started from A0 = {0}
contains, modulo boundary corrections, six copies of the
dynamics on Wn , it follows that
|A2n −1 | = 3bn − 3 · 2n + 1.
(3–1)
Next, consider the dynamics on the {0, . . . , L} ×
{0, . . . , L} box with initial occupied sites consisting of
the frame {(x, y) : x ∈ {0, L} or y ∈ {0, L}}. We call
L the size of such a hole. Holes of sizes 10 and 34 are
featured in Figure 2. Let H(L) be the final occupation
count minus the frame,
H(L) = |A∞ ∩ {(x, y) : 0 < x < L, 0 < y < L}|.
We define
hn = H(2n + 2).
Thus h0 = 2, h1 = 6, h2 = 20, etc. Actually, it is slightly
more convenient to use hn = hn −2n+1 . By observing the
wedge dynamics from time 2n to time 2n+1 , we obtain
our first basic recursion:
wn+1 = 3wn + 2(wn−1 − 2n−1 ) + hn−1 .
This recursion is illustrated by the top row of Figure 2,
where the five smaller wedges are outlined. Here one
thinks of the sites in Z2 as centers of the squares, and
the outlined regions include their boundaries. The sites
outside the outlined regions correspond to the interior of
the pictured size-10 hole, from which the bottom row and
rightmost column are removed. Therefore
bn+1 = 3bn + 2bn−1 + 2hn−1 .
(3–2)
A second basic recursion is obtained by observing the
hole dynamics until time 2n−1 :
hn = 2bn−1 + 2hn−1 .
This is illustrated in the bottom row of Figure 2, where
the interiors of the first descendant holes are outlined.
Equivalently,
hn = 2bn−1 + 2hn−1 .
(3–3)
Express hn in terms of b’s using (3–2), and then plug
into (3–3) to obtain the second-order equation bn+2 =
5bn+1 − 4bn , valid for n ≥ 1. Hence bn = α · 4n + β.
Computing the constants α, β from b1 and b2 , we get
bn =
5 n 2
·4 + .
6
3
429
Thus, starting from A0 = {0}, by (3–1),
|A2n −1 | =
5 n
· 4 − 3 · 2n + 3,
2
and ρ13 = 56 .
Hex 135 , Hex 136 , and Hex 1356 all satisfy precisely
the same recursions (3–2) and (3–3). Also, it is easy to
check that for small holes and wedges, Hex 13 and Hex
135 evolve identically. Thus, it follows by induction that
Hex 13 and Hex 135 crystals agree exactly for all t when
A0 = {0}. In particular, ρ135 = ρ13 = 56 . Since Hex
136 and Hex 1356 both fill small holes and wedges, it
also easily follows that these two rules solidify completely
from a singleton.
We turn next to Hex 134 , another exactly solvable rule
with density less than 1. The domain counts bn and wn
are defined in exactly the same manner as before, giving
b0 = 1, b1 = 4, b2 = 16, b3 = 62, b4 = 246, etc.
As mentioned in Section 2, the hole interactions (see
Figure 3) are more complicated now. There are three different types of hole dynamics, all run in the {0, . . . , L} ×
{0, . . . , L} box. We define the frame F = {(x, y) :
x ∈ {0, L} or y ∈ {0, L}}. All occupation numbers are
counted strictly inside this frame, and in addition exclude any other initially occupied sites. We will always
use L = 2n + 2 and count occupation numbers at time 2n
(which in every case differ by 1 from the final occupation
numbers).
The principal hole dynamics use initial occupied sites
consisting of the “frame minus 3 corner sites,” F \{(0, L−
1), (0, L), (1, L)}, and defines the occupation numbers hn .
Then h0 = 3, h1 = 7, h2 = 22, etc. Again, we set
hn = hn − 2n+1 .
The secondary hole dynamics start with two opposite
corners missing: F \ {(0, L − 1), (0, L), (1, L), (L − 1, 0),
(L, 0), (L, 1)}. Call the resulting occupation numbers sn .
Then s0 = 2, s1 = 6, s2 = 20, etc.
The third and last hole dynamics start with a triangular hole, i.e., the initially occupied set
(F \ {(0, L − 1), (0, L), (1, L)}) ∪ {(x, y) : y < x},
and define tn . Then t0 = 2, t1 = 5, t2 = 13, etc.
Our system of recursions is given by (3–2), again illustrated in the top frames of Figure 3, and
hn = 2bn−1 + wn−1 + tn−1 + sn−1 ,
sn = 2bn−1 + 2sn−1 ,
tn = 2wn−1 + hn−1 .
To understand these new equations, consult the bottom
row of Figure 3. The first successor holes (outlined) are a
430
FIGURE 3. Hex 134 : Basic wedge dynamics (top) at times 24, 28, and 32; final configuration of a hole of size 10;
size-34-hole dynamics at times 16, 24, and 28.
triangular one and a square one of secondary type. The
former creates a hole of the principal type, while the latter creates two holes of the secondary type, as seen in the
bottom middle frame. Therefore, we have
4.
DENSITY OF HEX 1 AND ITS COUSINS
In this section we will analyze, in this order, Hex 1, 14,
145, 15, 16 , and 156 . We still use the definition,
ρ = lim
5
7
= bn−1 + tn−1 + sn−1 − 2n ,
2
4
sn = 2bn−1 + 2sn−1 ,
3
tn = bn−1 + hn−1 + 2n .
2
n→∞
hn
(3–4)
At this point the problem could be solved by matrix manipulation, but it is easier to eliminate sn and tn using
the first and third equations of (3–4), then eliminate hn
using (3–2). This yields the equation bn+3 − 5bn−2 +
3bn−1 + 4bn = 0. It follows that bn is a linear combina√
tion of 4n , φn , and (−φ−1 )n , where φ = (1 + 5)/2 is
the golden ratio. Computing constants, the final result
is, for n ≥ 1,
√ n
√ 1+ 5
21 n 15 − 5
·4 +
·
bn =
22
55
2
√ n
√ 1− 5
15 + 5
·
+
.
55
2
|A2n −1 |
,
3 · 4n
where A0 = {0}. However, for these rules we cannot give
the exact value of the density; instead, we will demonstrate that the limit exists. It is, for now, convenient to
redefine
bn
ρ = lim n ;
n→∞ 4
we prove that (4–1) also holds at the end of Section 6.
The numbers bn are defined by appropriate wedge dynamics, defined slightly differently than before. Namely,
these dynamics run inside the first quadrant with one
boundary condition on the axes, and with the initial occupied set A0 consisting of the single point (apart from
the axes) (1, 2). Now set
Bn = {(x, y) : 1 ≤ x ≤ 2n , 1 ≤ y ≤ 2n + 2}
and let bn be the size of the final occupied set inside Bn ,
i.e.,
bn = |A∞ ∩ Bn |.
We also make use of the smaller wedge
Wn = {(x, y) : 1 ≤ x ≤ 2n , 2 ≤ y ≤ x + 1},
21
22 .
Again (3–1) holds, so ρ134 =
Once more, it is easy to prove that from A0 = {0},
Hex 1345 generates exactly the same A∞ as Hex 134 ,
while Hex 1346 and Hex 13456 solidify completely.
(4–1)
and define
wn = |A∞ ∩ Wn |.
Note that bn = 2wn − 1 − (2n − 1) = 2wn − 2n .
431
FIGURE 4. Hex 1 : Basic wedge dynamics (top) at times 18, 24, and 32; final configuration of hole of size 10; size-34-hole
dynamics at times 17, 26, and 32.
Appropriate hole dynamics on the {0, . . . , L} ×
{0, . . . , L} box have initial occupied sites consisting of
the frame {(x, y): x ∈ {0, L} or y ∈ {0, L}} together
with (1, 2) and (L − 2, L − 1). As before,
Our second recursion is generated by hole dynamics
run until growth from the initial two buds collides:
H(L) = |A∞ ∩ {(x, y) : 0 < x < L, 0 < y < L}|,
The restriction on n, obtained from Lemma 4.1 below,
is not optimal, but is one that works in all cases. (For
small k one can get away with a less-restrictive condition,
which varies from case to case and is useful for computations.) The error terms ek , which keep this recursion
from closing, are the result of slightly “dirty” interaction between the two growing buds (cf. the bottom of
Figure 4).
We bound ek using the following fact about additive
dynamics. Starting from T0 = {0}, perform 2n − 1 steps
to generate T2n −1 . Fix an , 0 < ≤ 2n , and let Z =
Z(n, ) be the union of all connected components (in the
triangular lattice sense) of 0’s in T2n −1 ∩ {(x, y) : y ≥
2n − }.
but now we need a much larger variety of hole counts:
hkn = H(2n − 2k),
k = −1, 0, 1, . . . .
Here we view k as fixed and n large enough that this
makes sense. We also abbreviate hn = h−1
n .
Two basic recursions will be derived in a manner similar to the Hex 13 analysis. The only difference between
the six rules is the correction term in (4–2) below. Equation (4–3) is the same in all cases.
We now proceed with Hex 1 .
The first recursion is obtained by observing Wn from
time 2n to 2n+1 :
wn+1 = 3wn + 2(wn−1 − 2n−1 ) + hn−1 − 2(2n−1 − 1)
= 3wn + 2wn−1 + hn−1 − 2n+1 + 2.
Note (as in the top left frame of Figure 4) that two rows
and columns are now removed from the size-10 hole to
match the sites not covered by the five wedges. Therefore
bn+1 = 3bn + 2bn−1 + 2hn−1 − 2n+1 + 4.
(4–2)
For example, b0 = 1, b1 = 4, b2 = 14, b3 = 50, b4 =
182, . . . ; and h0 = 0, h1 = 2, h2 = 8, h3 = 36, h4 =
154, . . . .
hkn = 2bn−1 + hkn−1 + hk+1
n−1 − ek ,
2n−1 ≥ 4k + 8. (4–3)
Lemma 4.1. Every (x, y) ∈ Z has y ≥ 2n − 2.
Proof: Focus on the line {y = 2n − } and consider an
interval of a 0’s flanked by 1’s at both ends. Find the first
1 below, say at distance b, the leftmost 0 of this interval.
Paint this occupied site red . Then the column of b + 1
sites (b 0’s and the red site) must have to its immediate
left a column of b+1 1’s. The dynamics now ensures that
a = b and that the red site is connected by an occupied
diagonal to the 1 at the right border of the initial interval
of 0’s.
432
This proves that the worst case is that in which T2n −1 ∩
{y = 2n − } consists of − 1 0’s flanked by two 1’s, in
which case b = − 1.
One can in principle compute ek for any k from quantih
w
ties ew
k and ek , which we now define. First, ek is the final
occupation count in the region at the tip of the growth
that gives bn , consisting of a (2k + 3) × (2k + 3) box together with two lattice triangles, a (2k + 3) × (2k + 3) and
a (2k + 1) × (2k + 1) one. Here is the region for k = 1,
labeled with x’s:
1
1
1
1
x
x x
x x x
x x x x
1
1
x
x
x
x
x
1
x
x
x
x
x
1
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
1
1
1
ehk
is the final count of occupied sites in the
Next,
interaction area in the middle of H(2n − 2k). This area
consists of a (2k+3)×(2k+3) box and four triangles—the
two above and their reflections across the main diagonal
of the box. (The set of occupied sites is symmetric with
respect to this reflection at all times.)
In all small cases we use, we have checked that the last
occupied site in the interaction area (or in fact anywhere
outside the two smaller holes that are filled recursively)
gets added at time 2n−1 . If this is the case, a speedof-light argument shows that ew
n can be computed in the
(2k+3)×(2k+3) box with an additional row of 2k+3 sites
at the bottom, while the interaction area for computation
of ew
k adds to this an additional column of 2k + 3 sites on
the right. Then (4–3) holds under the restriction 2n−1 ≥
2k + 6. As mentioned earlier, we use this property of
small cases for computations.
In any case, by Lemma 4.1 above (and its proof),
h
ek = 2ew
k − ek .
Direct enumeration gives e−1 = 2, e0 = 10, e1 = 24,
e2 = 44, e3 = 66, e4 = 92, e5 = 128, e6 = 170, e7 = 212,
e8 = 258.
Observe that if we know hK
n , n ≤ N , and ek , k ≤ K−1,
then we can use (4–2) and (4–3) to compute bn up to
n ≤ N + K + 3.
An explicit formula for ek is apparently too much to
hope for, but using the trivial facts that ehk and ew
k are
nonnegative and bounded above by the number of sites
in their respective regions, we do have the bounds
2
−3(2k + 3)2 < −ehk ≤ ek ≤ 2ew
k < 4(2k + 3) .
Write
η(n) =
n−1
k k=0 i=0
k
(2i + 1)2 = n2 2n − n2n + 2n − 1.
i
Using (4–3) repeatedly, and the above bounds, one obtains
n
hn =
2k bn−k − en ,
k=1
where
−3η(n) ≤ en ≤ 4η(n).
(4–4)
Therefore
bn+1 = 3bn + 2bn−1 + 2
n−1
2k bn−1−k − en−1 − 2n+1 + 4
k=1
= 3bn +
n
2k bn−k − en ,
(4–5)
k=1
where en = en−1 + 2n+1 − 4 for n ≥ 1 and e0 = −1.
The sequence bn /4n satisfies a renewal equation, and
the renewal theorem [Feller 68, p. 330] gives
∞
∞
1 − 14 n=0 4−n en
bn
5 1 −n −
=
4 en .
ρ1 = lim n = 3 1 ∞
−k
n→∞ 4
6 6 n=1
k=2 k2
4 + 2
(4–6)
Summability of 4−n en guarantees existence of the limit,
but ρ1 could be 0 or 1 in principle.
To obtain nontrivial bounds on ρ1 , we enumerate h9n
for n ≤ 11 directly, employ (4–2) and (4–3) to compute
bn exactly for n ≤ 23, and then use (4–5) to compute en
for n ≤ 22. For n ≥ 23, we estimate en as follows.
The lack of symmetry between the upper and lower
bounds in (4–4) can be removed by counting 0’s instead
of 1’s. That is, write bcn = 2n (2n + 2) − bn and hcn =
(2n + 1)2 − hn . Then the analogues of (4–2) and (4–3)
lead to
bcn+1 = 3bcn +
n
2k bcn−k − en−1 − 4 · 2n − 6.
(4–7)
k=1
Although these en are different from the previous ones,
we do not introduce new notation, because they satisfy
the same bounds (4–4). Changing (4–7) into an equation
for bn , and using the result to obtain the upper bound
for en in (4–5), we get
−3η(n−1)+2n+1 −4 ≤ en ≤ 3η(n−1)+2n·2n −3·2n −6.
(4–8)
The dominant term in both upper and lower bounds is
3n2 2n . This can in fact be improved to 52 n2 2n , since a
better bound for ek is obtained by considering where the
two wedge dynamics and the hole dynamics must agree.
A little geometric argument demonstrates that this agreement is achieved at least within a region between lines
of slope 2 and 12 through the center of the interaction
region. We omit the details, since it is much easier to
improve the bounds on ρ1 by computing more ek ’s for
small k than by trying to improve (4–8).
Our computations yield 4−23 b23 ≈ 0.635280, and so,
using (4–6) and (4–8), the rigorous bounds
0.635248 ≤ ρ1 ≤ 0.635312.
Our second rule is Hex 14 . With minor changes our
analysis also applies to Hex 145 , so we will omit that
case.
The first basic recursion now reads, for n ≥ 3,
wn+1 = 3wn + 2wn−1 + hn−1 − 2 · 2n − 2,
and this time
bn = 2wn + 2n .
This yields, with corrections computed separately for
low n,
bn+1 = 3bn +2bn−1 +2hn−1 −6·2n −4−1{n=1} +2·1{n=2} ,
(4–9)
for n ≥ 1. We have b0 = 3, b1 = 8, b2 = 24, b3 = 78,
b4 = 280, . . . , and h0 = 4, h1 = 8, h2 = 25, h3 = 80, . . . .
The second recursion remains (4–3).
What does change in this case is computational.
Namely, in all small cases we use for our estimates, the
computation for bn ends at time 2n + 1, and at time
2n−1 + 1 in the interaction area. For computation of ehk
this forces us to add two layers of 2k + 3 sites at the bottom and on the right of the central (2k + 3) × (2k + 3)
box and one layer of 2k + 5 sites at the top and on the
left of that box. For computation of ew
k , the added layers
are of course only at the bottom and on the left, creating
a (2k + 5) × (2k + 4) box. (The restriction in (4–3) then
is 2n−1 ≥ 2k + 8.) In this way, we get e−1 = 2, e0 = 11,
e1 = 27, e2 = 49, e3 = 82, e4 = 125, and e5 = 170.
Thus
n
bn+1 = 3bn +
2k bn−k − en ,
k=1
where en = en−1 + 6 · 2n + 4 + 1{n=1} − 2 · 1{n=2} for
n ≥ 1, e0 = 1, and en satisfies the same bounds (4–4).
In this case the renewal theorem gives
∞
∞
3 − 14 n=0 4−n en
bn
11 1 −n −
ρ14 = lim n = 3 1 ∞
=
4 en .
−k
n→∞ 4
6 6 n=1
k=2 k2
4 + 2
433
The estimates of en for large n are as follows (this time
we do not bother to symmetrize):
−3η(n − 1) + 6 · 2n + 4 ≤ en ≤ 4η(n − 1) + 6 · 2n + 4.
For this case we compute bn up to n = 20, to get
4−20 b20 ≈ 0.968854 and
0.968618 < ρ14 < 0.969044.
Next, we turn to Hex 16 . All definitions, as well as
recursions (4–2) and (4–3), remain exactly the same as
for Hex 1 . Note that the “6” part of the rule influences
only sites that have no further influence elsewhere, so
this rule has exactly the same interactions as Hex 1 . For
example, we can obtain A∞ for Hex 1 and then perform
a single Hex 16 step to obtain A∞ for Hex 16 .
The extra step does affect the computation of ek , making in small cases the interaction areas the same as for
the Hex 14 rule. To repeat: for ehk we add two layers of
2k + 3 sites at the bottom and on the right of the central (2k + 3) × (2k + 3) box and one layer of 2k + 5 sites
at the top and on the left of that box, while for ew
k the
added layers are only at the bottom and on the left. The
restriction in (4–3) is 2n−1 ≥ 2k + 8.
We get b0 = 1, b1 = 4, b2 = 16, b3 = 58, b4 = 212,
etc., and e−1 = 4, e0 = 15, e1 = 32, e2 = 56, e3 = 83,
e4 = 115, and e5 = 157. By computing h6n up to n = 9,
we obtain 2−18 b18 ≈ 0.739664 and
0.738902 < ρ16 < 0.740279.
Next in line is Hex 156 . The first recursion now is
similar to (4–9),
bn+1 = 3bn +2bn−1 +2hn−1 −6·2n +10−6·1{n=1} , (4–10)
while (4–3) is still the second recursion. In addition,
bn = 2n (2n + 2) up to n = 3; after that, permanently
empty triangles (of six sites) appear. Much later (see
the discussion in Section 8), larger permanently empty
regions appear.
In this case, two extra steps are required (again, for the
small cases we have checked) to finalize the configuration
in Bn . The last of these two steps, however, merely fills
two sites next to the axes and thus does not affect any
computations. The interaction areas are therefore the
same as for Hex 16 .
This time, we get b0 = 3, b1 = 8, b2 = 24, b3 = 80,
b4 = 276, etc., and e−1 = 1, e0 = 9, e1 = 25, e2 = 55,
e3 = 84, e4 = 106, and e5 = 151. By computing h6n up
to n = 9, we obtain 2−18 b18 ≈ 0.937935 and
0.937183 < ρ16 < 0.938559.
434
Our last rule in this section is Hex 15 . In this case,
the first recursion is
bn+1 = 3bn +2bn−1 +2hn−1 −6·2n +8−6·1{n=1} , (4–11)
and we get b0 = 3, b1 = 8, b2 = 22, b3 = 70, b4 = 238,
etc., and e−1 = 0, e0 = 6, e1 = 18, e2 = 42, e3 = 66,
e4 = 84, and e5 = 120. By computing h6n up to n = 9,
we obtain 2−18 b18 ≈ 0.802578 and
0.801822 < ρ16 < 0.803199.
We note that again small cases require two steps to finalize the configuration in Bn . Both steps fill two sites
next to the axes and are thus not problematic. The new
feature is different hole dynamics, requiring one extra
step (beyond 2n−1 ) to resolve. Therefore, the interaction
areas are the same as for Hex 16 .
5.
DENSITY OF HEX 1456 AND HEX 146
For concreteness, we will concentrate on Hex 1456 . The
same techniques applied to Hex 146 yield the bounds in
Theorem 1.2. The definitions of bn and hn are the same
as in Section 4. The first recursion now is
bn+1 = 3bn + 2bn−1 + 2hn−1 − 6 · 2n − 2,
(5–1)
while the second is still (4–3). We note that if hn ≡
(2n + 1)2 , then bn ≡ 2n (2n + 2). That is, if every hole
fills completely, then An fills the lattice and ρ1456 = 1.
At first, simulations suggest that this indeed happens.
However, after more systematic experimentation one
discovers that H(82) and H(84) do not fill completely;
a final triangle of 0’s consists of 25 sites. For the lower
bound, it is also important to observe that all the other
holes of sizes at most 130 do fill in. Therefore, all boxes
of sizes 2n − 44, n ≥ 7, do not fill in. (In fact, they leave
2(n − 7) + 1 unfilled triangles of size 25.) Since there are
infinitely many such boxes generated by the dynamics,
the number of unfilled 0’s is infinite. This, however, does
not establish that ρ1456 < 1. In fact, we do not even have
a nontrivial lower bound for the density yet, so we start
with this easier task.
A zero-creating hole will occur for the first time inside
an H(2n + 2) when 2n−22 − 44 ≥ 84; hence n ≥ 28. Thus
bn = 2n (2n + 2) for n ≤ 29. (This explains why a naive
simulation started with a single occupied site will never
produce a hole: one would need a system of a size more
than 230 ≈ 109 .)
As in (4–7), let bcn = 2n (2n + 2) − bn and hcn = (2n +
1) − hn . Then
2
bcn+1 = 3bcn + 2bcn−1 + 2hcn−1
and hcn satisfy the analogue of (4–3), with the same
bounds (4–4) on ek . Hence
bcn+1 = 3bcn +
n
2k bcn−k − en−1 ,
(5–2)
k=1
where en = 0 for n ≤ 28. Furthermore, en satisfies the
bounds (4–4). The renewal theorem therefore guarantees
the existence of a density, and, by (4–4),
1 − ρ1456
∞
bcn
1 −n = lim n = −
4 en
n→∞ 4
24 n=29
<
∞
1 −n 2 n
282
4 ·n 2 <
< 5.3 · 10−7 .
8 n=29
8 ln 2 · 228
Unfortunately, a nontrivial upper bound seems very
difficult to produce by a brute-force approach along the
lines of Section 4. Instead, we devise a rescaling argument.
Let R be a time at which there is a permanent 0 in AR
(with the initial condition that produces bn ). Clearly we
can choose R to be a power of 2, and arbitrarily large. We
can also ensure that the 0 is at distance at least R/2 (in
· ∞ distance, say) away from the part of the boundary
of AR strictly inside the first quadrant.
Now run the dynamics in time steps of size R, with
the proviso that whenever a hole is formed, the clock
inside this hole is reset so that time 0 corresponds to the
formation of the hole. The dynamics then uses different
clocks on different regions of the plane. The point is
that every second time, starting with time 1, every added
occupied site is part of a version of AR (its translation
and/or rotation). The other times are used for filling in,
and holes cannot be guaranteed.
As we know, the buds inside the holes interact, but
the interaction will not reach the existing 0 until it is of
size R/2, therefore up to (unrescaled) time at least 2R/4 .
Up to that time, therefore, all the occupied sites will see
a permanent 0 within distance 4R.
Next, let dn be the site count of sites within distance
4R of a permanent 0, at time 2n +1. Since the interaction
can destroy a 0 and therefore affect sites at distance 4R
from the interaction area, dn satisfy the same recursion
as bn , but with error bounds multiplied by 81 · R2 . If
dcn = 2n (2n + 2) − dn , then
dcn+1 = 3dcn +
n
k=1
2k dcn−k − en−1 ,
(5–3)
435
where in this version
|en | < 103 R2 n2 2n .
Moreover, we already know that en = 0 for n ≤ R/4.
Therefore
lim sup
dcn
<
103 R2 n2 2−n ,
n
4
n≥R/2
which clearly can be made arbitrarily small for large
enough R.
It follows that a positive proportion of occupied sites
sees a permanent 0 within a fixed distance. Therefore
the final density of 1’s is necessarily strictly less than 1.
6.
PROOF OF THEOREM 1.1
We start with the case A0 = {0}. What we need to
prove, roughly, is that the proportion of A∞ -sites in any
nice large subset of Z2 is about ρ. We accomplish this
by the division of space into “mesoscopic” units filled in
by the basic wedge dynamics (which gives wn ) and much
smaller problematic regions.
To this end, let us return to the recursion for wn .
This recursion can be “unrolled.” That is, when not
both π(3) and π(4) are 1, wn can be written in terms
of wn−m−1 , . . . , wn−m− , and hkn−m− , for some m ≥ 0,
≥ 2, and suitable k’s. This is the case we will focus on
for the remainder of the proof. When π(3) = π(4) = 1,
then hkn−m− is replaced by a suitable linear combination
of hn−m− , sn−m− , and tn−m− , and the proof is easily
adapted. (The right side of Figure 5 illustrates the difference between resulting divisions.) The unrolled recursion
corresponds to a division of Wn into smaller triangles and
parallelograms. If n is large, and Wn is normalized by
1/2n , this partition approximates a partition of the unit
triangle W = {(x, y) : 0 ≤ y ≤ x ≤ 1}. In this way, W
is divided into covering triangles (corresponding to wi )
and parallelograms. We define these sets to be closed,
so there is some intersection along boundaries. Here, m
determines the largest size of a triangle used, which is
1/2m+1 times the size of W . Furthermore, gives exactly the number of different triangle sizes used. The top
of Figure 5 pictures the division for m = 1 and = 3,
with the remaining parallelograms shaded. The maximum diameter of sets in the division clearly halves with
each successive m. Moreover, the area covered by the
parallelograms (uncovered by triangles) halves with each
successive . View m and as large, but much smaller
than n.
FIGURE 5. The division of W for m = 1 and = 3 into two
cases.
Fix an > 0. It is clear that |4−(n−m−i) wn−m−i − 12 ρ|,
i = 1, . . . , , can be made simultaneously smaller than if n is chosen large enough. (Note that 12 · 4−(n−m−i)
is exactly the area of the corresponding triangle in the
division.)
Pick a closed set S ⊂ W , with nice boundary. For
example, we may assume that S is convex and ∂S consists
of finitely many linear pieces. How many points from
2−n (A∞ ∩ Wn ) does S contain? In the sequel, C is a
“generic constant,” which is allowed to vary from one
appearance to another.
First, the triangles that intersect ∂S together contain
at most C · length(∂S) · 22n−m points. Second, all the
parallelograms in W have combined area at most C · 2− ,
and so they contain at most C · 22n− points. A crude
upper bound on the total length of the boundaries of the
division sets is 4 times the number of division sets, which
can be (equally crudely) bounded by the reciprocal of the
area of the smallest set in the division. Therefore, the
total boundary length is at most C · 4m+ , and so within
(microscopic) distance C of these boundaries there are
at most C · 2n+2m+2 points. Finally, we know that the
error terms (if any) contribute at most C · n2 · 2n points.
Therefore, S contains at least
(ρ − ) · area(S) − C · length(∂S) · 2−m − C · 2−
− C · 2−n+2m+2 − C · n2 · 2−n · 4n
points, for large enough n. An analogous upper bound
also holds. By letting first n → ∞, then m, → ∞, and
436
finally → 0, it follows that
1
|S ∩ 2−n (A∞ ∩ Wn )| → ρ · area(S).
4n
The proof of Theorem 1.1 for A0 = {0} is now concluded in a straightforward manner.
For general finite A0 , Theorem 1.1 mostly follows from
properties of additive dynamics, as we explain below.
The proof is somewhat delicate, which is not a complete
surprise, since sensitivity to perturbations in initial condition is widespread among generic cellular automaton
rules. In fact, there is a “box snowflake” for which the
density of A∞ has been proved to depend on the initial
set. Namely, we have shown in [Gravner and Griffeath 98]
that Box 1 solidification yields density 49 starting from
a singleton. Later, Dean Hickerson [Hickerson 99] engineered finite initial seeds with asymptotic densities 29
64
61
and 128
. For instance, the latter is achieved by an ingenious arrangement of 180 carefully placed occupied cells
around the boundary of an 83 × 83 grid. We do not know
the highest density with which Box 1 solidification can
fill the plane, nor whether any seed fills with density less
than 49 .
To return to hexagonal digital snowflakes, we will assume that A0 has sites on the y-axis, the lowest of which
is at 0 and the highest at (0, h), but no sites to the right
of the y-axis. We will concentrate on proving the density result for the part of A∞ between lines y = 0 and
y = x+h, which, by symmetry and the extreme boundary
dynamics, is clearly enough. We start with the following fact about additive dynamics. Assume the seed is
T0 = A0 ∩ y-axis. By additivity, when 2n − 1 > h, the
rightmost column of T2n consists of two copies of T0 , separated by 2n − 1 − h 0’s. The column immediately to the
left of these 0’s consists either of all 1’s (flanked by two
0’s) or else of all 0’s, flanked on the top by a 1 and on
the bottom by an additional 0, below which there is a 1.
In the second case we can iterate and continue moving to
the left until we encounter a column of 1’s (flanked by a
0 on both ends). Assume that this column consists of D
1’s. An important point is
2n − h ≤ D ≤ 2n .
Furthermore, if this column is created (by the boundary
dynamics) at time t, then at time t + D + 1 the boundary
dynamics creates a horizontal and a diagonal segment
that together with the above column seal off a triangle of
0’s. Into this triangle (which is a bounded perturbation
of the one started from a singleton), the Hex dynamics
spread as when they fill basic wedges.
However, interaction between the secondary boundary
dynamics is slightly more complicated. When the boundary dynamics from the two secondary buds (cf. Section
2) collide, they (together with the primary boundaries)
create a hole that is less than or equal in size to that from
A0 = {0}. In fact, the hole’s side can be diminished by
at most h − 1. By Lemma 4.1, the hole generates smaller
holes (all of which are smaller, and bounded perturbations of the A0 = {0} case), provided that all the wedge
vertices that were dead before are still dead . The problematic vertices are created in the middle of the holes
when secondary boundary dynamics collide. If π(3) and
π(4) are not both 1, collisions always result in dead vertices, as is easy to check. We will deal with this case
first.
Pick a closed convex set S ⊂ W with piecewise linear
boundary and an > 0. In the first step, perform m
steps of the Sierpiński triangle construction. That is,
take out of W the central triangle congruent to W/2,
then three triangles congruent to W/4, . . . , and finally
3m triangles congruent to W/2m . Discard the points in
W outside these triangles (see top left of Figure 6 for the
m = 3 example). In the second step, mark by M the two
smaller triangles from each triangle obtained in the first
step, as in Figure 6. In the third step, mark four triangles
in the remaining quadrilateral hole, then four triangles in
each of the remaining two holes, and so on, for a total
of m iterations and 4 · 2m triangles. The top of Figure 6
illustrates these three steps for m = 3. Note that when
m = m() is large enough, the area of the exceptional set,
i.e., the set of points outside of marked triangles, is below
/4. For a small enough δ = δ(, m), the area of We , the
δ-neighborhood of the exceptional set, is below /2.
Now take a large n and consider 2−n A∞ ∩ W on one
of the marked triangles WM , off We . By Lemma 2.1, this
configuration exactly equals the configuration of sites in
a basic wedge if n ≥ n0 (δ). Therefore, by taking n even
larger if necessary,
|2−n A∞ ∩WM ∩Wec ∩S| ≥ ρ·area(WM ∩Wec ∩S)−6−m−1 ,
by the already proved result for basic wedges. Since the
number of marked triangles is less than 2 · 6m , it follows
that
|2−n A∞ ∩ S| ≥ ρ · area(S) − ,
for n ≥ n0 (), as desired. The analogous upper bound is
proved in the same fashion.
When π(3) = π(4) = 1, the middle vertices in the
holes are live as soon as D < 2n . Although h > 0 does
not necessarily imply that D < 2n , there is for each h a
FIGURE 6. The three steps and marked triangles for m = 3.
Regions outside them are shaded.
unique T0 that has D = 2n , namely the one generated
by the additive dynamics starting from {0} and run for
h time steps (and translated back so that its lowest site
is at 0).
In this case, therefore, the division into marked triangles in step 3 has two possible forms, one for D = 2n
and one for D < 2n . Both are indicated in the bottom
of Figure 6. In either case, the proof is then a minor
modification of the above.
This concludes the proof of Theorem 1.1. To finish
this section, we assume π(3) = 0 and prove (4–1), or
equivalently, that for the dynamics on Wn ,
4−n |A2n −1 ∩ Wn | →
1
ρ.
2
In fact, the recursion for wn can be used to show that
A2n +Cn2 ∩ Wn = A∞ ∩ Wn , since the largest interaction
area inside the holes is on the order of n2 . (It is possible
that in all cases, A2n +c ∩ Wn = A∞ ∩ Wn , for some
small constant c, but we cannot prove this.) Then, in
the division as in the case A0 = {0} above, only the
dynamics inside triangles that intersect the right edge
of W is not done by time 2n . These contain at most
C · 22n−m points, and the proof is concluded as before.
(Growth inside some parallelograms might also persist,
but such sites are already incorporated into the error.)
437
There are two kinds of macroscopic dynamics for digital snowflakes, both denoted here by Sa , a ∈ [0, 1]. The
sets Sa will be a strictly increasing family of closed subsets of the hexagon co(N ) ⊂ R2 . We start with the
simpler case, which we call simple hole dynamics (SHD).
This evolution will be associated with the 12 rules for
which π(3) and π(4) are not both 1.
Assume a < 1 and write the dyadic expansion a =
0.a1 a2 . . . with infinitely many 0’s. We set a = 0.a2 a3 . . .,
a = (a ) , etc., and use the following transformations:
reflection σ about the line y = x, rotation ρα through an
angle α, and deformation δ by the linear transformation
(x, y) → (x − y, y). We will define only the portion of Sa
inside the triangle W = {(x, y) ∈ R2 : 0 ≤ x ≤ y ≤ 1};
other parts are then obtained by symmetry, i.e., by an
appropriate use of the above transformations.
We will first recursively define Sa ⊂ W , and a corresponding hole dynamics Ha ⊂ W on dyadic rationals a
that have only finitely many 1’s in their dyadic expansion, starting with Sa = {(0, 0)} and Ha = ∅ for a = 0.
The rules described in the next paragraph and Figure 7
are a natural extension of the recursions for wn and hn
for the corresponding 12 rules. (We remark that when
no two successive ai ’s are 1, one can define the boundary
of Sa by a Koch-type substitution scheme, as outlined in
Section 1 for a = 13 and in [Gravner and Griffeath 98]
for Box 1 solidification; such algorithms are in general
precluded by collisions inside the holes.)
If a1 = 0, then take Sa = 12 Sa . If a1 = 1 and a2 = 0,
then Sa consists of five pieces:
Sa = 12 W ∪ 12 , 0 + 12 Sa ∪ 12 , 12 + 12 Sa
∪ 12 , 0 + 14 σSa ∪ 12 , 12 + 14 ρ−π/2 δSa .
If a1 = 1 and a2 = 1, then Sa consists of seven pieces:
Sa = 12 W ∪ 12 , 0 + 12 Sa ∪ 12 , 0 + 12 Sa
∪ 12 , 12 + 14 σW ∪ 12 , 12 + 14 ρ−π/2 δW
∪ 12 , 14 + 14 Ha ∪ 1, 12 + 14 ρπ Ha .
If a1 = 0, then
Ha = (1, 1) + 12 ρπ σSa ∪ (1, 0) + 12 ρπ/2 δSa ,
and if a1 = 1, then
7.
MACROSCOPIC DYNAMICS
In this section we assume A0 = {0} and study At for a
large finite t instead of A∞ . As the radius of At increases
linearly in t, it is natural to ask whether these dynamics
have a hydrodynamic limit as space and time are scaled
proportionally. Computer simulations certainly suggest
so, and are validated by our analysis below.
Ha = (1, 1) + 12 ρπ σW ∪ (1, 0) + 12 ρπ/2 δW ∪ 12 Ha
∪ 1, 12 + 12 ρπ Ha .
It is clear from the construction that both Sa and Ha
are well defined and increasing on dyadic rationals. Moreover, we claim that |a − b| < 2−n implies
√
√
dH (Sa , Sb ) ≤ 2 · 2−n , dH (Ha , Hb ) ≤ 2 · 2−n , (7–1)
438
a1 = 0
a1a2 = 10
Sa
Sa
a1a2 = 11
Sa
Ha
Sa
Sa :
a1 = 0
a1 = 1
Sa
Ha
Ha :
FIGURE 7. SHD. Dark areas are fully covered by appropriately mapped W , while arrows indicate the placement of
suitably transformed sets.
where dH is the Hausdorff metric. To prove (7–1), we
proceed by induction on n. Note that we can assume
a < b and, by monotonicity, that ak = 0, for k ≥ n and
bn = 1. The assumption also means that ak =
√ bk for
k < n. Since diameter(W ) = diameter(δW ) = 2, it is
easy to verify (7–1) for n = 0 and n = 1. For n ≥ 2,
we can use the induction hypothesis on a , a and b , b .
Note that all a sets are normalized by 12 and all a sets
by 14 , which yields (7–1).
The uniform continuity (7–1) immediately implies
that we can extend both Sa and Ha uniquely to a ∈ [0, 1],
so that (7–1) holds for a, b ∈ [0, 1], S1 = H1 = W , and
FIGURE 8. Occupied set of Hex 13 , started at {0}, at times
, and 62
in darker, lighter, and darker
a · 210 for a = 23 , 14
15
63
shade, respectively. This illustrates problems with surmising
fractal properties from pictures: the 23 case appears to have
the “most fractal” boundary, which in fact has dimension 1.
The fairly large fractal dimension in the other two cases is a
consequence of growth that appears at the exposed corners
and is not visible at this scale.
dimension dimH of the said boundary. The answer in
many cases is yes, and we turn to this task now.
As in [Gravner and Griffeath 98], we need to count the
number of buds of size 2−n . Let us denote this number
by fn (a) for the triangle and gn (a) for the (half) hole
dynamics. Then
Sa = ∩b>a Sb = closure(∪b<a Sb ), Ha = ∩b>a Hb
⎧
⎪
if a1 = 0,
⎨fn−1 (a )
fn (a) = 2fn−1 (a ) + 2fn−2 (a ) if a1 a2 = 10,
⎪
⎩
2fn−1 (a ) + 2gn−2 (a ) if a1 a2 = 11,
= closure(∪b<a Hb ),
where b in all cases ranges over dyadic rationals. The following result identifies Sa as a subsequential limit of At .
and
Theorem 7.1. Assume that tn = a · 2n , for a ∈ [0, 1] and
A0 = {0}. The 12 rules that do not have both π(3) = 1
and π(4) = 1 exhibit subsequential convergence to SHD:
2−n Atn → Sa ,
in the Hausdorff metric, as n → ∞, uniformly in a.3
Proof: For dyadic rationals a, the proof follows from arguments in Section 6. The case of general a ∈ [0, 1] is
then a consequence of monotonicity of At and St .
The sets Sa are difficult to visualize for a’s that are not
dyadic rationals. In particular, it is clear that they have
a very complicated boundary. It is therefore natural to
ask whether one can determine the fractal (Hausdorff)
3 The limit also equals ρS in the weak sense, but Hausdorff
a
convergence seems more to the point.
2fn−1 (a )
gn (a) =
2gn−1 (a )
if a1 = 0,
if a1 = 1.
When a is rational, these equations reduce to a finite linear recursion, so that fn (a) ∼ cλn , for some c > 0 and
λ ∈ (0, 4]. If λ ≤ 2, then it follows immediately that
dimH (∂Sa ) = 1. If λ > 2, then these are Mauldin–
Williams [Mauldin and Williams 88] fractals. Hence
the box-counting and Hausdorff dimensions agree, and
dimH (∂Sa ) = log λ/ log 2.
To illustrate, for N ≥ 2 let us take a with periodic
binary representation that repeats N − 1 initial 1’s and
then a single 0, so a = (2N − 2)/(2N − 1). Index the
shifted configurations with superscripts. We get
N −1
2
3
= 4gn−2
= · · · = 2N −2 gn−(N
gn1 = 2gn−1
−2)
= 2N −1 fn−(N −1) .
a1a2 = 10
a1 = 0
Sa
Sa
Sa
a1a2 = 11
Ha3
Ha1
Sa
Sa :
a1 = 0
a1 = 1
Ha2
Sa
439
a1 = 0 or a1 a2 = 10, the recursion for Sa is the same as
before, while if a1 a2 = 11, then
Sa = 12 W ∪ 12 , 0 + 12 Sa ∪ 12 , 0 + 12 Sa ∪ 12 , 0 + 14 W
∪ 12 , 12 + 14 ρ−π/2 δW ∪ 12 , 14 + 14 Ha1
∪ 1, 12 + 14 ρπ Ha3 .
If a1 = 0, then
Ha1 :
Ha1 = 12 Sa ∪ (1, 1) + 12 ρπ σSa ∪ (1, 0) + 12 ρπ/2 δSa ,
a1 = 0
Ha2 = Ha3 = (1, 1) + 12 ρπ σSa ∪ (1, 0) + 12 ρπ/2 δSa ,
a1 = 1
Sa
Ha1
while if a1 = 1,
Ha3
Ha1 = 12 W ∪ (1, 1) + 12 ρπ σW ∪ (1, 0) + 12 ρπ/2 δW
Ha2 :
∪ 12 ρπ Ha2 ,
a1 = 0
Ha2 = (1, 1) + 12 ρπ σW ∪ (1, 0) + 12 ρπ/2 δW ∪ 12 ρπ Ha1
∪ 1, 12 + 12 ρπ Ha3
a1 = 1
Sa
Ha3
Ha3 :
FIGURE 9. DHD. Live vertices within the H’s are indicated
by arrows.
Also,
N
fnN −1 = fn−1
= fn−1 ,
N −1
N
+ 2fn−2
= 4fn−2 ,
fnN −2 = 2fn−1
N −2
N −1
+ 2gn−2
= 12fn−3 .
fnN −3 = 2fn−1
Recursively thereafter we get
N −(k−1)
fnN −k = 2fn−1
fnN −k
N −(k−2)
+ 2gn−2
These are extended to a ∈ [0, 1] in the same way as
before, yielding the following result.
Theorem 7.2. Assume that tn = a · 2n , for a ∈ [0, 1], and
A0 = {0}. The four rules with π(3) = 1 and π(4) = 1
exhibit convergence to DHD:
2−n Atn → Sa ,
in the Hausdorff metric as n → ∞, uniformly in a.
= ck fn−k , using
= (2ck−1 + 2k−1 )fn−k .
Therefore c1 = 1, and for k ≥ 1, ck satisfy the recursion
ck = 2ck−1 + 2k−1 . It follows that ck = k · 2k−1 , so
fN = N · 2N −1 fn−N , and λ = 2 · (N/2)1/N . Hence
dimH (∂Sa ) = 1 −
Ha3 = (1, 1) + 12 ρπ σW ∪ (1, 0) + 12 ρπ/2 δW ∪ 12 ρπ Ha3
∪ 1, 12 + 12 ρπ Ha3 .
log2 N
1
+
.
N
N
For example, dimH (∂S2/3 ) = 1 and dimH (∂S14/15 ) =
dimH (∂S245/255 ) = 54 ; the dimension achieves its maximum (within this collection of examples) of approximately 1.289 at a = 62
63 , and tends to 1 as N → ∞.
See Figure 8.
Note that we do not know how to determine dimH (Sa )
when a is irrational, nor the maximum possible value
of dimH (Sa ).
The diverse hole dynamics (DHD) is associated with
the four Hex rules that have π(3) = π(4) = 1. Besides Sa ,
there are now three hole dynamics Hai ⊂ W , i = 1, 2, 3,
all initialized at ∅ when a = 0 (cf. Figure 9). When
It follows from Section 6 that for these four rules, most
finite seeds give rise to macroscopic dynamics different
from DHD. The more general construction is quite analogous, however, so we will not describe it in detail.
Dimension computations are similar to those for SHD,
but more complicated. Now, there are fn (a) and gni (a),
i = 1, 2, 3, that satisfy
⎧
if a1 = 0,
⎪
⎪fn−1 (a )
⎪
⎨2f
(a
)
+
2f
(a
)
if a1 a2 = 10,
n−1
n−2
fn (a) =
1
⎪2fn−1 (a ) + gn−2 (a )
⎪
⎪
⎩
3
(a )
if a1 a2 = 11,
+ gn−2
and
3fn−1 (a ) if a1 = 0,
gn1 (a) =
2
(a )
if a1 = 1,
gn−1
2fn−1 (a )
if a1 = 0,
gn2 (a) =
1
3
gn−1 (a ) + gn−1 (a ) if a1 = 1,
2fn−1 (a ) if a1 = 0,
gn3 (a) =
3
(a ) if a1 = 1.
2gn−1
440
FIGURE 10. Dynamics in H(84) at times 34, 49, 56, 58, 66. Arrows indicate which directions of boundary dynamics (from
nearby buds) lead to the residual hole.
For the same example a = (2N −2)/(2N −1) as in the SHD
case, a similar but more involved computation yields
dimH (∂Sa )
=
8.
1
· log2
N
1
7
11 5
N · 2N + 2N −
− 1{N
3
9
9
9
even}
.
two buds: b1 at (2, 2n−1 +4) (moving eastward) and b2 at
(2n−1 +2, 2n−1 +4) (moving diagonally toward the northwest), and their two symmetrically located counterparts
b1 and b2 . Now let us pause at time t2 = t1 + 2n−2 − 1
(49 in the example). The two buds b2 and b2 would have
created two diagonals
(2n−1 + 2 − 2n−2 + 1, 2n−1 + 4)
THICKNESS AND RELATED ISSUES
Let us begin the proof of Theorem 1.3 by considering the
eight digital snowflakes that are not exactly solvable. For
the argument that A∞ is never thick, we address only the
hardest case, Hex 1456 ; the same method applies equally
well to the other seven rules.
The key fact (initially suggested by simulations) is
that holes of sizes in the sequence below, obtained by
doubling and subtracting 4,
84, 164, 324, 644, . . . ,
create holes, each of which is twice as large as the previous one. Although this should follow from a general
rescaling property of the dynamics (which we do not even
know how to formulate), the property is a consequence
of a finite chain of basic interactions, as illustrated by a
sequence of intermediate times for H(84) in Figure 10.
Before we write out the details, note that the configuration in the hole is symmetric with respect to the map
(x, y) → (L − y, L − x), and write L = 2n + 2n−2 + 4,
n ≥ 6. The first important configuration to consider
occurs at time t1 = 2n−1 + 2 (t1 = 34 when n = 6). Concentrate on the extreme boundary dynamics generated by
to (2n−1 + 2, 2n−1 + 4 + 2n−2 − 1)
and
(L − (2n−1 + 4) − 2n−2 + 1, L − (2n−1 + 2))
to (L − (2n−1 + 4), L − (2n−1 + 2) + 2n−2 − 1)
if not for the interaction between them. By additivity,
we need to eliminate the intersection between the two,
which consists of two sites, as is easy to check. At the
same time, b1 creates a column
(2n−2 + 2, 2n−1 + 4) to (2n−2 + 2, 2n−1 + 4 + 2n−2 − 1).
Note that this does not interfere with the diagonal
created by b2 (on which the smallest x-coordinate is
2n−2 + 3).
Now it is easy to see that the gap of two sites created
by the interaction between b2 and b2 creates two buds
that generate the same boundary dynamics (in the northwest direction) as would a single 1 at time t2 − 1 immediately southeast from the gap, at (2n−1 + 2, 2n−1 + 2n−2
+2). Call this seed b3 . Also consider the bud b4 (and its
symmetric counterpart b4 ) at (2n−2 + 3, 2n−1 + 2n−2 + 4)
0
0
0
0
0
02
1
01
0
0
01
0x
FIGURE 11. Local configuration at time t − 1.
(immediately northeast from the top of the column in
the previous paragraph), which creates extreme boundary dynamics moving eastward. The next time to pause
is just before these two interact, at time t3 = t2 +2n−3 −1
(which is 56 in our example). At this time, b3 creates a
diagonal
(2n−1 + 2 − (2n−3 − 1), 2n−1 + 2n−2 + 2)
to (2n−1 + 2, 2n−1 + 2n−2 + 2 + 2n−3 − 1)
plus two new buds at its ends, one to the west of the
bottom point, one to the north of the top point. It follows that the boundary dynamics from b4 is free to make
one more step free from interruption from b3 to create a
column
(2n−1 − 2n−3 + 2, 2n−1 + 2n−2 + 4)
to (2n−1 − 2n−3 + 2, 2n−1 + 2n−2 + 4 + 2n−3 − 3).
This column would be higher if not for interference from
b4 , which creates a horizontal column at the same time.
These two, together with an additional site from b3 , seal
the hole at time t3 + 1. Three new sites are occupied
inside at the next time, but the remaining sites remain 0
forever (if n ≥ 6).
Since the dynamics create a hole of every size in the
sequence, infinitely many times and no matter what the
finite seed A0 (cf. Sections 4 and 6), it follows that there
are arbitrarily large islands of 0’s in the final state A∞ .
Therefore A∞ is not thick.
The argument that Ac∞ is always thick for Hex 1 is
quite simple. Namely, we will show that every 1 in A∞
that is at ∞ -distance 3 or more from A0 must have a 0
within ∞ -distance 2. Assuming that this is not the case,
let a 1 at x be a counterexample. Let t be the time at
which x becomes 1. We label 0x the 0 at x at the time
t − 1. We will also assume, without loss of generality,
that the single 1 in x + N at this time is to the left of x.
It follows that the two 0’s labeled 01 in Figure 11 must
also become 1 at time t (to avoid being a 0 with two
neighboring 1’s), and then so does 02 . This forces the
remaining 0’s in Figure 11 and yields a contradiction,
since the 1 cannot be isolated from other 1’s.
Let us turn next to the exactly solvable rules. To show
that A∞ is thick when A0 = {0}, we simply note that
441
the smallest wedges and holes contain 1’s, and choose
a suitable R, an upper bound on the distance between
any 0 and the set of 1’s. An easy induction shows that
R provides the same bound for larger wedges and holes.
General initial sets are handled by arguments in Section
6. (Note, however, that the maximal distance of a 0 from
the set of 1’s in A∞ may not be bounded independently
of A0 .) Thickness of Ac∞ is equally easy to establish.
To finish the proof of Theorem 1.3, we need to demonstrate point (2).
We start with Hex 13456 . First note that a finite seed
will generate six rays of 1’s via the extreme boundary
dynamics. Assume that there is a nonempty connected
(in the hexagonal lattice sense) set of 0’s in A∞ . If this
set is infinite, then it must extend to infinity, say between
the eastern and northeastern ray of 1’s. From now on we
refer to the sites within that portion of A∞ . Start at some
point between the rays, and move up to the topmost 0;
call it 01 . Assume first that its northeast neighbor is
0; call it 02 . There cannot be any 0’s above 02 (or else
the topmost 0 would see at least three 1’s). Then the
northeast neighbor of 02 must be a 0, call it 03 , again
with no 0’s above it.
Continuing in this fashion, we create an infinitely extended diagonal of 0’s. But the boundary dynamics alone
precludes this, for otherwise they would never create a 1
directly above the diagonal, and then directly above that,
etc., finally eliminating the ray of 1’s. Therefore the right
neighbor of 01 must be a 0, again call it 02 , and the entire
column above 02 must be 1’s. By the same argument, the
right neighbor of 02 , named 03 , must be a 0 and the entire
column above it 1’s. This procedure eventually creates a
horizontal ray of 0’s, which is equally impossible by the
boundary dynamics. The only remaining possibility is
that the connected set of 0’s is finite. But then simply
observe that the rightmost among its top 0’s must see
at least three 1’s. This contradiction demonstrates that
Ac∞ = ∅, as claimed.
We conclude this section by showing that Hex 13456
is the only case for which A∞ does not have infinitely
many 0’s regardless of A0 . It is easy to observe that the
other three candidates (i.e., rules with density 1) may
leave some 0’s: Hex 136 and Hex 1346 will not fill a
domino 00 in a sea of 1’s, while Hex 1356 will not fill a
triangle 000 in a sea of 1’s.
To generate infinitely many 0’s, let the initial set A0
consist of a column of L 1’s. Then the primary boundary
encloses a triangle with this column as one side, at time
L. This repeats every time the boundary dynamics recreates the initial set, hence infinitely many times. To
442
show that there is an A0 with infinitely many 0’s in A∞ ,
it is therefore enough to find an L for which the described
triangle does not completely fill in.
By exhaustive computer search, the smallest such L’s
for rules Hex 136 , Hex 1346 , and Hex 1356 are, respectively, 5, 19, and 42. We do not know whether there are
smaller seeds with this property, but it seems that very
small random initial seeds (those that fit into a 5×5 box,
say) are very likely to have A∞ = Z2 .
9.
EXACT SOLVABILITY
The phrase “exact solvability” is popular in statistical
physics [Baxter 82], but its exact meaning is difficult to
pin down. By contrast, as we will now explain, in the discrete world of deterministic CA this concept does have a
natural rigorous framework based on computational complexity. We will restrict our formulations to the final set
A∞ , although generalization to the entire space-time evolution is straightforward. (In fact, at the cost of an extra
dimension, it is easy to represent the evolution of any
two-state CA as the final configuration of a solidification
rule.) Intuitively, we require that “for a given x ∈ Z2 ,
it is computationally easy to decide whether x ∈ A∞ .”
This notion of course depends on an initial set A0 , which
will be assumed to be {0} unless otherwise specified. Formally, we call a solidification CA exactly solvable (from
A0 ) if there exists a finite automaton that upon encountering x as input, decides whether x ∈ A∞ . Representation of x as input is given as (±i11 , ±i21 , i12 , i22 , . . . ), where
i11 , i21 are the most significant binary digits of the first and
second coordinates of x; i12 , i22 the next-most significant,
etc. (Some initial i1k ’s or i2k ’s may be 0, and the representation is finite but of arbitrary length.) This means
that A∞ is automatic [Allouche and Shallit 03], or equivalently a uniform tag system [Cobham 72]. More general
representations of inputs are sometimes desirable, particularly if one wants to study CA on more general lattices,
but this one suffices for our present purposes. Note that
exact solvability puts a limit on the computational complexity of a CA: for example, it cannot be universal.4
Also, by [Cobham 72, Theorem 6] such A∞ cannot have
an irrational density, in contrast to the limit set in [Griffeath and Hickerson 03] generated by a suitable initial
seed in the game of life.
4 Note that since there are one-dimensional universal CA [Cook
05], there are two-dimensional universal solidification CA. A universal CA can generate any recursive language, hence one that is
not recognizable by a finite automaton.
To our knowledge, the simplest nontrivial example
of an exactly solvable CA is Diamond 1 solidification,
which is a modification of Hex 1 using the neighborhood
{(0, 0), (0, ±1), (±1, 0)}. In this case (the “intricate, if
very regular, pattern of growth” depicted on [Wolfram
02, p. 171]), it can be shown by induction that x ∈
/ A∞
iff max{k : i1k = 1} = max{k : i2k = 1}. It is easy to
construct a (two-state) finite automaton that checks this
condition, and the density ρ of A∞ evidently must satisfy
the equation ρ = 12 + ρ4 , so that ρ = 23 . It is also worth
noting that the first-quadrant portion of this A∞ is a
10
fixed point of the substitution system 1 → 10
11 , 0 → 01 .
Although, by Cobham’s theorem [Cobham 72, Theorem
3], [Allouche and Shallit 03, Theorem 14.2.3], a substitution representation of A∞ must exist in any exactly
solvable case, we have no simple explicit construction for
our Hex examples.
The proof that Hex 13 is exactly solvable is quite a bit
messier than the recursion for bn , since symmetry seems
difficult to exploit. We now sketch the construction of the
required finite automaton, as encoded in Figure 12. The
number of states will be determined by a number of possible division squares. Except in three cases, these squares
are further divided into two triangles by the northeast diagonal. Each of these triangles is either a wedge, in which
case it is equipped with an arrow at the vertex from which
the wedge propagates in the dynamics, or a half-hole that
is equipped with two arrows. The half-hole encodes dynamics filling in one-half of the diamond-shaped holes.
We need to distinguish wedges and half-holes with different orientations, so there are six of the former and four
of the latter. The first exceptional (undivided) division
square is a full hole, which has a single representative and
encodes filling a square hole. The other two exceptions,
arbitrarily named B-square and C-square, have four representatives each and encode filling the “complex” part
of a wedge. (A B-square is marked by three arrows, while
a C-square is marked by a single arrow, as shown in Figure 12.) The number of possible division squares can be
reduced by observing that two half-holes cannot appear,
and neither can two wedges with arrows that originate
at the corners of the southeast diagonal. This makes a
total of 29 possible division squares.
Assume that both initial signs are +. Draw the initial division square, representing the initial state of the
automaton, which has two wedges, each with an arrow
at the lower left corner. The situation is reflected when
both the initial signs are −, while in the case of a + and
a − the initial state is a C-square.
00
W
W
W
01
W
B
443
B
10
11
H
W
H
W
B
W
W
FIGURE 13. Hole correction. First correction (on the larger
scale) is in the direction (0, 1), next in the direction (0, −1).
B
W
C
W
W
W
C
W
C
W
F
W
W
F
One can also prove that Hex 134 is exactly solvable by
the same method. The algorithm is a little more involved,
since it necessitates several more types of full holes and
half-holes, but there is nothing conceptually new, so we
omit the details.
As it turns out, we have already established enough
structure to establish the relative complexity of the eight
remaining digital snowflakes. Indeed, the following result
[Gravner and Hickerson 06], together with Theorems 1.1–
1.3, proves that none of them can be exactly solvable from
any initial set.
W
F
W
W
W
F
H
H
H
W
FIGURE 12. Division algorithm. Transition according to
the value of i1 , i2 is indicated in the top line, and pieces are
labeled in the obvious fashion.
Lemma 9.1. If an automatic set S ⊂ Z2 has a strictly
positive asymptotic density, then it is thick.
In conclusion, we remark that complexity analysis of
cellular automata by means of finite automata was initiated by S. Wolfram in [Wolfram 84b]. The perspective
of Wolfram’s paper is in a sense opposite to ours, in that
it measures how the complexity of the set of all possible configurations increases over time. In contrast, we
consider a single configuration at the “end of time.”
ACKNOWLEDGMENTS
After reading (i11 , i21 ), the next state is one of the four
subsquares, two of the same type as the initial one, the
other two B-holes. This division is of course dictated by
how the dynamics fills the square. The next two bits
induce further division, and so forth. Figure 12 provides
a check that any of the 29 division squares gets divided
into 4 division squares.
A minor final complication is that the holes are not
divided into two smaller holes by exact wedges, but by
their translation in a coordinate direction by 1. These
corrections do not accumulate, since they go in opposite
directions from one scale to the next, as illustrated by
Figure 13.
We thank Ken Libbrecht for sharing with us his unmatched
expertise on real snowflakes, and Mirek Wójtowicz for developing the MCell cellular automata explorer. We are also
grateful to Dean Hickerson, who pointed out that the rules
Hex 14 , Hex 145 , Hex 146 , and Hex 1456 , started from a
singleton, all generate different final sets.
The first author was partially supported by NSF grant
DMS–0204376 and the Republic of Slovenia’s Ministry of Science program P1-285. The second author was partially supported by NSF grant DMS–0204018.
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Janko Gravner, Mathematics Department, University of California, Davis, CA 95616 ([email protected])
David Griffeath, Department of Mathematics, University of Wisconsin, Madison, WI 53706 (griff[email protected])
Received August 16, 2005; accepted in revised form March 26, 2006.

Modeling Snow Crystal Growth I: Rigorous Results

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